A projectile is shot from the ground at an angle of #(5 pi)/12 # and a speed of #7/15 m/s#. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

Question

1277 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-11T19:45:10

0.0115 m (assuming #g = 10 m/s^2#

The maximum height the projectile reaches is

#h = {u^2 sin^2 theta}/{2 g} =0.01016 m#

The horizontal distance it travels up to this point is

#d = {u^2 sin theta cos theta}/g = 0.00544 m#

The distance of this point from the starting point is

#sqrt{h^2+d^2} = 0.0115 m$